Hey Veeky Forums I got an easy question for you

Hey Veeky Forums I got an easy question for you.
No, it's not my homework, I've already solved it. I'll post the answer 41 minutes after this is posted.

Again, I'm not saying this question is hard, it's just simple but fun.

16

when the side of the small corner square is equal to 1/4 the side of the big square.

How did you figure that out?

u friggin sirius m8

Yes. Did you eyeball it? I stated it's not to scale, you need to prove it algebraically.

When the ones on the outside are 25% of the inside. Are you even trying?

jesus...
say the side of the small sq = a
and one in the middle is 2b
then for the equality:
4a^2 +4b^2 = 8ab
a^2 -2ab +b^2 =0
(a-b)^2 =0
since length cannot be negative a=b
one side of the big sq = 2a+2b =4a
etc...

>dat image
every. single. time.

Well let the side of the big square be s. Then the side of the small square is s/4.

The gray area is:

4 (s/4)^2 + (s - 2*s/4)^2 = (s^2)/ 4 + (s^2)/4 = 1/2 s^2

The rectangles in white have dimensions s/4 by (s - 2*s/4) = s/2. They have area 1/8 s^2 . Now there are 4 of them. So the white area is s^2/2.

You can also start out with the the assumption that the area of the big square is is s^2, and try to find the size of the small square that makes the grey or white area (s^2)/2, since in order for the two to be equal they must be half the area of the big square.

It's honestly so trivial that "by inspection" would probably be accepted.

>cut into 4
>ask yourself when grey = white, remembering they're squares
>obviously when the two greys are equal

Wrong - you cannot know that the small side is s/4 right away

Correct

Area of gray: (a-2b)^2+4b^2
Area of white: 4(a-2b)*b
(a=side of large square, b=side of small gray squares)
Equating,
[math] b=-\frac{1}{16} (a-8)a [/math]

The problem only ask you to show that it's true, not to derive it.

Crap, sorry inputted into wolfram wrong, it should be
[math] b=\frac{a}{4} [/math]

Did you miss the big bold red text saying "not to scale"? Sure it's trivial but it's just a simple math exercise, not asking you to solve an unsolvable problem.

What the fuck do you think "show that it's true" means?

small grey = a^2
big grey = b^2
white rectangle = ab

4a^2 + b^2 = 4ab

4a^2 -4ab + b^2 = 0

(b-2a)^2 = 0

b-2a = 0

b = 2a

a : Small grey square (area of all of them is 4a^2)

b: Center grey square (area of it is b^2)

white square area = 4ab

4a^2+b^2=4ab
b=2a

>algebraically
>by inspection

I did show that it's true though. I showed that if the side of the small rectangle is 1/4th the size of big rectangle, then the areas are equal.

*sighs*

[math]A_{grey} = A_{white}[/math]
[math]a^2 + b^2 =2ab[/math]
[math]a = b[/math]

Brainlets, I swear.

>*sighs*
>only person in the thread to fuck it up

that methodology (induction) will almost never work elsewhere, where it isn't so easy to make educated guesses

Correct

Incorrect

Actually it's how most of science is conducted since the times of Francis Bacon and David Hume and before.

that's pretty much why science took baby steps until descartes/cartesian/deductive reasoning and then boomed after that

Did you miss the part where I used the symmetry of the problem to cut it into 4?

>brainlets

let [math]x[/math] be the length of the sides of the larger grey square, and [math]y[/math] be the length of the sides of the smaller grey squares.
As the white rectangle shares two sides with smaller grey squares and one with the larger square, the area must be equal to [math]xy[/math].
Given there are 4 white rectangles, 4 smaller grey squares and one larger grey square; under the circumstances that the white and grey areas are equal
[eqn]x^2+4y^2=4xy[/eqn]
[eqn]x^2-4xy+y^2=0[/eqn]
[eqn](x-2y)^2=0[/eqn]
[eqn]x-2y=0[/eqn]
[eqn]x=2y[/eqn]
Hence the grey area is equal to the white area when the sides of the larger square are double that of the smaller square.

I get the feeling you don't know very much about science and the history of science.

>substituting the problem and forgetting to undo the substitution before giving an answer
>everyone is stupid but me for incomplete thoughts

ah, shite, missed the 4 off the [math]y^2[/math] on the second equation, but the rest is correct.

nice try spidey

Length of corner grey squares = x
Length of large centre grey square = y

[math]
A_w = 4xy\\

A_g = 4x^2 + y^2\\

A_g = A_w\\

y^2-4xy+4x^2=0\\
(y-2x)^2=0\\
y=2x
[/math]

big square edge twice the small edge

more easily done by just cutting the whole square into 4x4 grid

The gray ad white aren't necessary equal, though. Is it asking to find proportions (line lengths) so that the two areas are equal?

This is the answer. Originally I set it up with L being the side of the large square, and s the side of each of the small, corner squares. Therefore, L - 2 is the length of a side of the center square. Then it turned into a cluster fuck. This is the easiest way to do it for sure.